Two point charges of $1 \, \mu\text{C}$ and $5 \, \mu\text{C}$ are placed at a distance of $4 \, \text{cm}$ apart. The ratio of the forces exerted by them on each other will be:

Why is the Coulombian force called a two-body force?

Two point charges $+8 \mu C$ and $+12 \mu C$ repel each other with a force of $48 \ N$. When an additional charge of $-10 \mu C$ is given to each of these charges (the distance between the charges is unaltered),then the new force is:

Two small spheres,each having an equal positive charge $Q$ (Coulomb),are suspended by two insulating strings of equal length $L$ (metre) from a rigid hook. The whole setup is taken into a satellite where there is no gravity. The two balls are now held by electrostatic forces in a horizontal position. The tension in each string is then:

Two point charges placed in air separated by distance $R$ exert a force $F$ on each other. Now the space between those charges is filled with a medium of dielectric constant $K$. At distance $R_1$ between the charges, the same force is exerted as $F$. Then:

Three point charges of magnitude $5 \mu C$,$0.16 \mu C$,and $0.3 \mu C$ are located at the vertices $A$,$B$,and $C$ of a right-angled triangle whose sides are $AB = 3 \, cm$,$BC = 3 \sqrt{2} \, cm$,and $CA = 3 \, cm$. Point $A$ is the right-angle corner. Calculate the magnitude of the net electrostatic force (in $N$) experienced by the charge at point $A$ due to the other two charges.